Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\cos 2x = 1 - 2{\sin ^2}x \Rightarrow {\sin ^2}x = \dfrac{{1 - \cos 2x}}{2}\\
{\sin ^2}A + {\sin ^2}B - {\sin ^2}C\\
= \dfrac{{1 - \cos 2A}}{2} + \dfrac{{1 - \cos 2B}}{2} - {\sin ^2}C\\
= 1 - \dfrac{1}{2}\left( {\cos 2A + \cos 2B} \right) - {\sin ^2}C\\
= \left( {1 - {{\sin }^2}C} \right) - \dfrac{1}{2}.2.\cos \dfrac{{2A + 2B}}{2}.\cos \dfrac{{2A - 2B}}{2}\\
= {\cos ^2}C - \cos \left( {A + B} \right).\cos \left( {A - B} \right)\\
= {\cos ^2}C + \cos \left( {180^\circ - A - B} \right).\cos \left( {A - B} \right)\\
= {\cos ^2}C + \cos C.\cos \left( {A - B} \right)\\
= \cos C.\left[ {\cos C + \cos \left( {A - B} \right)} \right]\\
= \cos C.2.\cos \dfrac{{C + A - B}}{2}.cos\dfrac{{C - A + B}}{2}\\
= 2\cos C.sin\left( {90^\circ - \dfrac{{C + A - B}}{2}} \right).sin\left( {90^\circ - \dfrac{{C - A + B}}{2}} \right)\\
= 2\cos C.\sin \dfrac{{\left( {A + B + C} \right) - \left( {C + A - B} \right)}}{2}.\sin \dfrac{{\left( {A + B + C} \right) - \left( {C - A - B} \right)}}{2}\\
= 2\cos C.\sin B.\sin A\\
= 2\sin A.\sin B.\cos C
\end{array}\)
